Partial pressures and the equilibrium constant KpAQA A-Level Chemistry: Revision notes
Section 1
Mole fraction and partial pressure
In a mixture of gases, the mole fraction of a gas is its amount divided by the total amount of gas:
mole fraction of A = moles of A ÷ total moles of gas
The partial pressure of a gas is the share of the total pressure that it contributes:
partial pressure of A = mole fraction of A × total pressure
The partial pressures of all the gases add up to the total pressure, and the mole fractions add up to 1.
Worked example: a mixture contains 0.30 mol of A and 0.70 mol of B at a total pressure of 100 kPa. The mole fraction of A is 0.30 ÷ 1.00 = 0.30, so p(A) = 30 kPa and p(B) = 70 kPa.
Always calculate the total moles at equilibrium first. A common error is to use the initial amounts rather than the amounts present at equilibrium.
Section 2
The expression for Kp
Kp is the equilibrium constant for a homogeneous gas-phase equilibrium at constant temperature, written using partial pressures. For aA(g) + bB(g) ⇌ cC(g) + dD(g):
Kp = p(C)ᶜ × p(D)ᵈ ÷ (p(A)ᵃ × p(B)ᵇ)
Products go on the top and reactants on the bottom, and each partial pressure is raised to the power of its coefficient. For N₂ + 3H₂ ⇌ 2NH₃, Kp = p(NH₃)² ÷ (p(N₂) × p(H₂)³).
The units of Kp are worked out from the expression. For the example above, kPa² ÷ (kPa × kPa³) = kPa⁻². For an equation with the same number of moles of gas on each side, Kp has no units.
Kp depends only on temperature. It does not change with pressure, concentration or a catalyst.
Do not write Kp with concentrations in square brackets. Kp uses partial pressures, written p(X), and the units follow from the expression.
Section 3
Calculations involving Kp
- Work out the amount (mol) of each gas at equilibrium.
- Find the total moles, then each mole fraction.
- Calculate each partial pressure = mole fraction × total pressure.
- Write the Kp expression, substitute and calculate, with units.
Worked example: 1.00 mol N₂ and 3.00 mol H₂ reach equilibrium, with 0.90 mol N₂, 2.70 mol H₂ and 0.20 mol NH₃ present, at a total pressure of 2000 kPa. Total = 3.80 mol. p(N₂) = 0.90 ÷ 3.80 × 2000 = 474 kPa, p(H₂) = 1421 kPa, p(NH₃) = 105 kPa. Kp = 105² ÷ (474 × 1421³) = 8.2 × 10⁻⁹ kPa⁻².
If Kp and all but one partial pressure are known, rearrange the expression to find the unknown. If only the reactant is given, the partial pressures of products formed in equal amounts are equal.
If you are told the amount of one substance at equilibrium, use the equation to work out the changes in the others. Check that the total moles at equilibrium makes sense.
Section 4
Effects of temperature, pressure and a catalyst
Pressure: increasing the total pressure shifts the position of equilibrium towards the side with fewer moles of gas. Kp is unchanged, because it only depends on temperature.
Temperature: this changes both the position and Kp. For an exothermic forward reaction, raising the temperature shifts the position left and Kp decreases. For an endothermic forward reaction, raising the temperature shifts the position right and Kp increases.
Catalyst: it speeds up the forward and reverse reactions equally, so equilibrium is reached more quickly. It does not change the position of equilibrium or the value of Kp.
In industry, a compromise is often needed: a lower temperature gives a better equilibrium yield for an exothermic reaction, but the rate is too slow.
A change in pressure can change the position of equilibrium but never the value of Kp. Only a change in temperature changes Kp.
Must know
- Partial pressure = mole fraction × total pressure; mole fraction = moles of gas ÷ total moles
- Kp = products' partial pressures (to the power of coefficients) ÷ reactants'
- The units of Kp come from the expression
- Pressure affects the position, not Kp; temperature affects both
- Exothermic: Kp decreases as temperature rises; endothermic: Kp increases
- A catalyst does not change Kp or the position, only the time to reach equilibrium
That's the notes covered.
Carry on to the next subtopic.
Exam questions on Partial pressures and the equilibrium constant Kp
- Dinitrogen tetroxide decomposes reversibly: N₂O₄(g) ⇌ 2NO₂(g). A sample of 1.00 mol of N₂O₄ is allowed to reach equilibrium in a sealed vessel at constant temperature. At equilibrium the mixture contains 0.60 mol of N₂O₄ and 0.80 mol of NO₂ and the total pressure is 200 kPa.Calculate the partial pressure of each gas in the equilibrium mixture.2 marks
- Ammonia is made in the Haber process: N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = −92 kJ mol⁻¹. An iron catalyst is used.State and explain the effect of the iron catalyst on the value of Kp and on the position of equilibrium.2 marks
- In a sealed vessel, 1.00 mol of phosphorus(V) chloride is heated at constant temperature until equilibrium is reached: PCl₅(g) ⇌ PCl₃(g) + Cl₂(g). The equilibrium mixture contains 0.40 mol of PCl₅, 0.60 mol of PCl₃ and 0.60 mol of Cl₂, and the total pressure is 150 kPa.Calculate the value of Kp for this equilibrium, and state its units.3 marks
Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).